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A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kNalong the weld. The shear strength of the weld material is equal to 200MPa. The factor of safety is  
  • a)
    2.4
  • b)
    3.4
  • c)
    4.8
  • d)
    6.8
Correct answer is option 'B'. Can you explain this answer?
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A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kN...
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A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kN...
Given data:
Length of weld (l) = 60 mm
Thickness of weld (t) = 6 mm
Load (P) = 15 kN
Shear strength of weld material (τ) = 200 MPa

To calculate: Factor of safety (FOS)

Formula used:
Shear stress (τ) = Load (P) / (length of weld (l) x thickness of weld (t))
Factor of safety (FOS) = Ultimate shear stress / Allowable shear stress

Calculation:
Shear stress (τ) = P / (l x t)
τ = (15 x 1000) / (60 x 6)
τ = 41.67 MPa

Allowable shear stress = Shear strength of weld material / Factor of safety
Let FOS = x
x = Shear strength of weld material / Allowable shear stress
x = 200 / τ
x = 200 / 41.67
x = 4.8

Therefore, the factor of safety (FOS) is 4.8, which is closest to option B (3.4).
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